Question

A body sliding on a frictionless track which terminates in a straight horizontal section CD. The body starts slipping from A and reaches point D. If h1 = 1 m, then how far from point D, will the body hit the ground?Take, g = 10 ms^-2.

• (A) 1 m
• (B) 2 m
• (C) 3 m
• (D) 4 m

Answer 1

The Correct Option is A

To solve this problem, we need to apply the principles of energy conservation and projectile motion.

Step-by-Step Solution

1. Energy Conservation:
   • The body starts at height h_1 = 1 \, \text{m} and slides down a frictionless track.
   • At the lowest point of the track, its potential energy is converted entirely into kinetic energy.
   • Applying conservation of energy between points A and D: mgh_1 = \frac{1}{2}mv^2 where v is the velocity at point D.
   • The potential energy at point D is mg\left(\frac{h_1}{2}\right), therefore: mgh_1 - mg\left(\frac{h_1}{2}\right) = \frac{1}{2}mv^2 which simplifies to: \frac{1}{2}mv^2 = \frac{1}{2}mgh_1
   • The velocity at point D is given by v = \sqrt{gh_1} and substituting g = 10 \, \text{m/s}^2, h_1 = 1 \, \text{m}: v = \sqrt{10 \times 1} = \sqrt{10} \, \text{m/s}
2. Projectile Motion:
   • From point D, the body travels horizontally. The height from which it falls is \frac{h_1}{2} = 0.5 \, \text{m}.
   • The time t to hit the ground is found using the equation of motion: s = \frac{1}{2}gt^2 where s = 0.5 \, \text{m}. Therefore: 0.5 = \frac{1}{2} \times 10 \times t^2 t^2 = 0.1 t = \sqrt{0.1} = 0.316 \, \text{s}
   • The horizontal distance x traveled is: x = vt = \sqrt{10} \times 0.316 \approx 1 \, \text{m}

Thus, the correct answer is (A) 1 m.